具有 return 类型的所有 class 方法名称的通用类型

Question

我正在尝试创建一个泛型，它将为我提供有关所有 class 方法 return 类型（+ 字段，但它们并不那么重要）的信息。

假设我创建了这样一个 class:

class TestClass {
    testField: string;

    constructor() {
        this.testField = 'test';
    }

    testMethod1(param: number): number {
        return param;
    }

    testMethod2(param: string): string {
        return param;
    }

}

结果，我想要这样的类型：

interface TestClassMethodsNamesReturnType {
  testMethod1: number;
  testMethod2: string;
}

到目前为止我尝试了什么

映射类型+类型推断：

示例 1

export type ClassMethodsNamesReturnType<T extends Function> = {
    [k in keyof T['prototype']]: ReturnType<T['prototype'][k]>;
}

const x: ClassMethodsNamesReturnType<TestClass> = {
    //...
}

错误：

Type 'TestClass' is missing the following properties from type 'Function': apply, call, bind, prototype, and 5 more

映射类型并将class视为对象：

示例 2

// I think here I can have problems to decide is the key of my class is a field or method 
// what can be a problematic to decide what I should use (typeof T[K] or ReturnType<typeof T[K]>)
export type ClassMethodsNamesReturnType<T extends Record<string, T[K]>, K extends keyof T> = {
    [K]: T[K];
}

错误：

A computed property name in a type literal must refer to an expression whose type is a literal type or a 'unique symbol' type

'K' only refers to a type, but is being used as a value here

你有什么提示或想法我怎样才能做到这一点？

Answer 1

虽然您需要使用映射类型中的 as 子句进行条件筛选，但您可以通过映射类型执行此操作。然后，检查该值是否扩展 Function 然后 return never 如果没有，那么它不会被映射。这是完整的代码：

class TestClass {
    testField: string;

    constructor() {
        this.testField = 'test';
    }

    testMethod1(param: number): number {
        return param;
    }

    testMethod2(param: string): string {
        return param;
    }

}

type FnReturns<T> = { [K in keyof T as T[K] extends Function ? K : never]: ReturnType<T[K] extends (...args: any[]) => any ? T[K] : never> };

// Correctly passes:
const foo: FnReturns<InstanceType<typeof TestClass>> = {
    testMethod1: 23,
    testMethod2: "hey",
}

// correctly fails:
const fail: FnReturns<InstanceType<typeof TestClass>> = {}

TypeScript Playground Link

另请注意我如何使用 typeof TestClass 的 InstanceType 来获取 TestClass.

的实例方法和属性

具有 return 类型的所有 class 方法名称的通用类型

Generic type for all class methods names with return type

generics

class

typescript

到目前为止我尝试了什么